optimization.py 31.7 KB
Newer Older
Stelios Karozis's avatar
Stelios Karozis committed
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862 863 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035 1036 1037 1038 1039 1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 1055 1056 1057 1058 1059 1060 1061 1062 1063 1064 1065 1066 1067 1068 1069 1070 1071 1072 1073 1074 1075 1076 1077 1078 1079 1080 1081 1082 1083 1084 1085 1086 1087 1088 1089 1090 1091 1092 1093
from copy import copy

from ..libmp.backend import xrange, print_

class OptimizationMethods(object):
    def __init__(ctx):
        pass

##############
# 1D-SOLVERS #
##############

class Newton:
    """
    1d-solver generating pairs of approximative root and error.

    Needs starting points x0 close to the root.

    Pro:

    * converges fast
    * sometimes more robust than secant with bad second starting point

    Contra:

    * converges slowly for multiple roots
    * needs first derivative
    * 2 function evaluations per iteration
    """
    maxsteps = 20

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        if len(x0) == 1:
            self.x0 = x0[0]
        else:
            raise ValueError('expected 1 starting point, got %i' % len(x0))
        self.f = f
        if not 'df' in kwargs:
            def df(x):
                return self.ctx.diff(f, x)
        else:
            df = kwargs['df']
        self.df = df

    def __iter__(self):
        f = self.f
        df = self.df
        x0 = self.x0
        while True:
            x1 = x0 - f(x0) / df(x0)
            error = abs(x1 - x0)
            x0 = x1
            yield (x1, error)

class Secant:
    """
    1d-solver generating pairs of approximative root and error.

    Needs starting points x0 and x1 close to the root.
    x1 defaults to x0 + 0.25.

    Pro:

    * converges fast

    Contra:

    * converges slowly for multiple roots
    """
    maxsteps = 30

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        if len(x0) == 1:
            self.x0 = x0[0]
            self.x1 = self.x0 + 0.25
        elif len(x0) == 2:
            self.x0 = x0[0]
            self.x1 = x0[1]
        else:
            raise ValueError('expected 1 or 2 starting points, got %i' % len(x0))
        self.f = f

    def __iter__(self):
        f = self.f
        x0 = self.x0
        x1 = self.x1
        f0 = f(x0)
        while True:
            f1 = f(x1)
            l = x1 - x0
            if not l:
                break
            s = (f1 - f0) / l
            if not s:
                break
            x0, x1 = x1, x1 - f1/s
            f0 = f1
            yield x1, abs(l)

class MNewton:
    """
    1d-solver generating pairs of approximative root and error.

    Needs starting point x0 close to the root.
    Uses modified Newton's method that converges fast regardless of the
    multiplicity of the root.

    Pro:

    * converges fast for multiple roots

    Contra:

    * needs first and second derivative of f
    * 3 function evaluations per iteration
    """
    maxsteps = 20

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        if not len(x0) == 1:
            raise ValueError('expected 1 starting point, got %i' % len(x0))
        self.x0 = x0[0]
        self.f = f
        if not 'df' in kwargs:
            def df(x):
                return self.ctx.diff(f, x)
        else:
            df = kwargs['df']
        self.df = df
        if not 'd2f' in kwargs:
            def d2f(x):
                return self.ctx.diff(df, x)
        else:
            d2f = kwargs['df']
        self.d2f = d2f

    def __iter__(self):
        x = self.x0
        f = self.f
        df = self.df
        d2f = self.d2f
        while True:
            prevx = x
            fx = f(x)
            if fx == 0:
                break
            dfx = df(x)
            d2fx = d2f(x)
            # x = x - F(x)/F'(x) with F(x) = f(x)/f'(x)
            x -= fx / (dfx - fx * d2fx / dfx)
            error = abs(x - prevx)
            yield x, error

class Halley:
    """
    1d-solver generating pairs of approximative root and error.

    Needs a starting point x0 close to the root.
    Uses Halley's method with cubic convergence rate.

    Pro:

    * converges even faster the Newton's method
    * useful when computing with *many* digits

    Contra:

    * needs first and second derivative of f
    * 3 function evaluations per iteration
    * converges slowly for multiple roots
    """

    maxsteps = 20

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        if not len(x0) == 1:
            raise ValueError('expected 1 starting point, got %i' % len(x0))
        self.x0 = x0[0]
        self.f = f
        if not 'df' in kwargs:
            def df(x):
                return self.ctx.diff(f, x)
        else:
            df = kwargs['df']
        self.df = df
        if not 'd2f' in kwargs:
            def d2f(x):
                return self.ctx.diff(df, x)
        else:
            d2f = kwargs['df']
        self.d2f = d2f

    def __iter__(self):
        x = self.x0
        f = self.f
        df = self.df
        d2f = self.d2f
        while True:
            prevx = x
            fx = f(x)
            dfx = df(x)
            d2fx = d2f(x)
            x -=  2*fx*dfx / (2*dfx**2 - fx*d2fx)
            error = abs(x - prevx)
            yield x, error

class Muller:
    """
    1d-solver generating pairs of approximative root and error.

    Needs starting points x0, x1 and x2 close to the root.
    x1 defaults to x0 + 0.25; x2 to x1 + 0.25.
    Uses Muller's method that converges towards complex roots.

    Pro:

    * converges fast (somewhat faster than secant)
    * can find complex roots

    Contra:

    * converges slowly for multiple roots
    * may have complex values for real starting points and real roots

    http://en.wikipedia.org/wiki/Muller's_method
    """
    maxsteps = 30

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        if len(x0) == 1:
            self.x0 = x0[0]
            self.x1 = self.x0 + 0.25
            self.x2 = self.x1 + 0.25
        elif len(x0) == 2:
            self.x0 = x0[0]
            self.x1 = x0[1]
            self.x2 = self.x1 + 0.25
        elif len(x0) == 3:
            self.x0 = x0[0]
            self.x1 = x0[1]
            self.x2 = x0[2]
        else:
            raise ValueError('expected 1, 2 or 3 starting points, got %i'
                             % len(x0))
        self.f = f
        self.verbose = kwargs['verbose']

    def __iter__(self):
        f = self.f
        x0 = self.x0
        x1 = self.x1
        x2 = self.x2
        fx0 = f(x0)
        fx1 = f(x1)
        fx2 = f(x2)
        while True:
            # TODO: maybe refactoring with function for divided differences
            # calculate divided differences
            fx2x1 = (fx1 - fx2) / (x1 - x2)
            fx2x0 = (fx0 - fx2) / (x0 - x2)
            fx1x0 = (fx0 - fx1) / (x0 - x1)
            w = fx2x1 + fx2x0 - fx1x0
            fx2x1x0 = (fx1x0 - fx2x1) / (x0 - x2)
            if w == 0 and fx2x1x0 == 0:
                if self.verbose:
                    print_('canceled with')
                    print_('x0 =', x0, ', x1 =', x1, 'and x2 =', x2)
                break
            x0 = x1
            fx0 = fx1
            x1 = x2
            fx1 = fx2
            # denominator should be as large as possible => choose sign
            r = self.ctx.sqrt(w**2 - 4*fx2*fx2x1x0)
            if abs(w - r) > abs(w + r):
                r = -r
            x2 -= 2*fx2 / (w + r)
            fx2 = f(x2)
            error = abs(x2 - x1)
            yield x2, error

# TODO: consider raising a ValueError when there's no sign change in a and b
class Bisection:
    """
    1d-solver generating pairs of approximative root and error.

    Uses bisection method to find a root of f in [a, b].
    Might fail for multiple roots (needs sign change).

    Pro:

    * robust and reliable

    Contra:

    * converges slowly
    * needs sign change
    """
    maxsteps = 100

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        if len(x0) != 2:
            raise ValueError('expected interval of 2 points, got %i' % len(x0))
        self.f = f
        self.a = x0[0]
        self.b = x0[1]

    def __iter__(self):
        f = self.f
        a = self.a
        b = self.b
        l = b - a
        fb = f(b)
        while True:
            m = self.ctx.ldexp(a + b, -1)
            fm = f(m)
            sign = fm * fb
            if sign < 0:
                a = m
            elif sign > 0:
                b = m
                fb = fm
            else:
                yield m, self.ctx.zero
            l /= 2
            yield (a + b)/2, abs(l)

def _getm(method):
    """
    Return a function to calculate m for Illinois-like methods.
    """
    if method == 'illinois':
        def getm(fz, fb):
            return 0.5
    elif method == 'pegasus':
        def getm(fz, fb):
            return fb/(fb + fz)
    elif method == 'anderson':
        def getm(fz, fb):
            m = 1 - fz/fb
            if m > 0:
                return m
            else:
                return 0.5
    else:
        raise ValueError("method '%s' not recognized" % method)
    return getm

class Illinois:
    """
    1d-solver generating pairs of approximative root and error.

    Uses Illinois method or similar to find a root of f in [a, b].
    Might fail for multiple roots (needs sign change).
    Combines bisect with secant (improved regula falsi).

    The only difference between the methods is the scaling factor m, which is
    used to ensure convergence (you can choose one using the 'method' keyword):

    Illinois method ('illinois'):
        m = 0.5

    Pegasus method ('pegasus'):
        m = fb/(fb + fz)

    Anderson-Bjoerk method ('anderson'):
        m = 1 - fz/fb if positive else 0.5

    Pro:

    * converges very fast

    Contra:

    * has problems with multiple roots
    * needs sign change
    """
    maxsteps = 30

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        if len(x0) != 2:
            raise ValueError('expected interval of 2 points, got %i' % len(x0))
        self.a = x0[0]
        self.b = x0[1]
        self.f = f
        self.tol = kwargs['tol']
        self.verbose = kwargs['verbose']
        self.method = kwargs.get('method', 'illinois')
        self.getm = _getm(self.method)
        if self.verbose:
            print_('using %s method' % self.method)

    def __iter__(self):
        method = self.method
        f = self.f
        a = self.a
        b = self.b
        fa = f(a)
        fb = f(b)
        m = None
        while True:
            l = b - a
            if l == 0:
                break
            s = (fb - fa) / l
            z = a - fa/s
            fz = f(z)
            if abs(fz) < self.tol:
                # TODO: better condition (when f is very flat)
                if self.verbose:
                    print_('canceled with z =', z)
                yield z, l
                break
            if fz * fb < 0: # root in [z, b]
                a = b
                fa = fb
                b = z
                fb = fz
            else: # root in [a, z]
                m = self.getm(fz, fb)
                b = z
                fb = fz
                fa = m*fa # scale down to ensure convergence
            if self.verbose and m and not method == 'illinois':
                print_('m:', m)
            yield (a + b)/2, abs(l)

def Pegasus(*args, **kwargs):
    """
    1d-solver generating pairs of approximative root and error.

    Uses Pegasus method to find a root of f in [a, b].
    Wrapper for illinois to use method='pegasus'.
    """
    kwargs['method'] = 'pegasus'
    return Illinois(*args, **kwargs)

def Anderson(*args, **kwargs):
    """
    1d-solver generating pairs of approximative root and error.

    Uses Anderson-Bjoerk method to find a root of f in [a, b].
    Wrapper for illinois to use method='pegasus'.
    """
    kwargs['method'] = 'anderson'
    return Illinois(*args, **kwargs)

# TODO: check whether it's possible to combine it with Illinois stuff
class Ridder:
    """
    1d-solver generating pairs of approximative root and error.

    Ridders' method to find a root of f in [a, b].
    Is told to perform as well as Brent's method while being simpler.

    Pro:

    * very fast
    * simpler than Brent's method

    Contra:

    * two function evaluations per step
    * has problems with multiple roots
    * needs sign change

    http://en.wikipedia.org/wiki/Ridders'_method
    """
    maxsteps = 30

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        self.f = f
        if len(x0) != 2:
            raise ValueError('expected interval of 2 points, got %i' % len(x0))
        self.x1 = x0[0]
        self.x2 = x0[1]
        self.verbose = kwargs['verbose']
        self.tol = kwargs['tol']

    def __iter__(self):
        ctx = self.ctx
        f = self.f
        x1 = self.x1
        fx1 = f(x1)
        x2 = self.x2
        fx2 = f(x2)
        while True:
            x3 = 0.5*(x1 + x2)
            fx3 = f(x3)
            x4 = x3 + (x3 - x1) * ctx.sign(fx1 - fx2) * fx3 / ctx.sqrt(fx3**2 - fx1*fx2)
            fx4 = f(x4)
            if abs(fx4) < self.tol:
                # TODO: better condition (when f is very flat)
                if self.verbose:
                    print_('canceled with f(x4) =', fx4)
                yield x4, abs(x1 - x2)
                break
            if fx4 * fx2 < 0: # root in [x4, x2]
                x1 = x4
                fx1 = fx4
            else: # root in [x1, x4]
                x2 = x4
                fx2 = fx4
            error = abs(x1 - x2)
            yield (x1 + x2)/2, error

class ANewton:
    """
    EXPERIMENTAL 1d-solver generating pairs of approximative root and error.

    Uses Newton's method modified to use Steffensens method when convergence is
    slow. (I.e. for multiple roots.)
    """
    maxsteps = 20

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        if not len(x0) == 1:
            raise ValueError('expected 1 starting point, got %i' % len(x0))
        self.x0 = x0[0]
        self.f = f
        if not 'df' in kwargs:
            def df(x):
                return self.ctx.diff(f, x)
        else:
            df = kwargs['df']
        self.df = df
        def phi(x):
            return x - f(x) / df(x)
        self.phi = phi
        self.verbose = kwargs['verbose']

    def __iter__(self):
        x0 = self.x0
        f = self.f
        df = self.df
        phi = self.phi
        error = 0
        counter = 0
        while True:
            prevx = x0
            try:
                x0 = phi(x0)
            except ZeroDivisionError:
                if self.verbose:
                    print_('ZeroDivisionError: canceled with x =', x0)
                break
            preverror = error
            error = abs(prevx - x0)
            # TODO: decide not to use convergence acceleration
            if error and abs(error - preverror) / error < 1:
                if self.verbose:
                    print_('converging slowly')
                counter += 1
            if counter >= 3:
                # accelerate convergence
                phi = steffensen(phi)
                counter = 0
                if self.verbose:
                    print_('accelerating convergence')
            yield x0, error

# TODO: add Brent

############################
# MULTIDIMENSIONAL SOLVERS #
############################

def jacobian(ctx, f, x):
    """
    Calculate the Jacobian matrix of a function at the point x0.

    This is the first derivative of a vectorial function:

        f : R^m -> R^n with m >= n
    """
    x = ctx.matrix(x)
    h = ctx.sqrt(ctx.eps)
    fx = ctx.matrix(f(*x))
    m = len(fx)
    n = len(x)
    J = ctx.matrix(m, n)
    for j in xrange(n):
        xj = x.copy()
        xj[j] += h
        Jj = (ctx.matrix(f(*xj)) - fx) / h
        for i in xrange(m):
            J[i,j] = Jj[i]
    return J

# TODO: test with user-specified jacobian matrix, support force_type
class MDNewton:
    """
    Find the root of a vector function numerically using Newton's method.

    f is a vector function representing a nonlinear equation system.

    x0 is the starting point close to the root.

    J is a function returning the Jacobian matrix for a point.

    Supports overdetermined systems.

    Use the 'norm' keyword to specify which norm to use. Defaults to max-norm.
    The function to calculate the Jacobian matrix can be given using the
    keyword 'J'. Otherwise it will be calculated numerically.

    Please note that this method converges only locally. Especially for high-
    dimensional systems it is not trivial to find a good starting point being
    close enough to the root.

    It is recommended to use a faster, low-precision solver from SciPy [1] or
    OpenOpt [2] to get an initial guess. Afterwards you can use this method for
    root-polishing to any precision.

    [1] http://scipy.org

    [2] http://openopt.org/Welcome
    """
    maxsteps = 10

    def __init__(self, ctx, f, x0, **kwargs):
        self.ctx = ctx
        self.f = f
        if isinstance(x0, (tuple, list)):
            x0 = ctx.matrix(x0)
        assert x0.cols == 1, 'need a vector'
        self.x0 = x0
        if 'J' in kwargs:
            self.J = kwargs['J']
        else:
            def J(*x):
                return ctx.jacobian(f, x)
            self.J = J
        self.norm = kwargs['norm']
        self.verbose = kwargs['verbose']

    def __iter__(self):
        f = self.f
        x0 = self.x0
        norm = self.norm
        J = self.J
        fx = self.ctx.matrix(f(*x0))
        fxnorm = norm(fx)
        cancel = False
        while not cancel:
            # get direction of descent
            fxn = -fx
            Jx = J(*x0)
            s = self.ctx.lu_solve(Jx, fxn)
            if self.verbose:
                print_('Jx:')
                print_(Jx)
                print_('s:', s)
            # damping step size TODO: better strategy (hard task)
            l = self.ctx.one
            x1 = x0 + s
            while True:
                if x1 == x0:
                    if self.verbose:
                        print_("canceled, won't get more excact")
                    cancel = True
                    break
                fx = self.ctx.matrix(f(*x1))
                newnorm = norm(fx)
                if newnorm < fxnorm:
                    # new x accepted
                    fxnorm = newnorm
                    x0 = x1
                    break
                l /= 2
                x1 = x0 + l*s
            yield (x0, fxnorm)

#############
# UTILITIES #
#############

str2solver = {'newton':Newton, 'secant':Secant, 'mnewton':MNewton,
              'halley':Halley, 'muller':Muller, 'bisect':Bisection,
              'illinois':Illinois, 'pegasus':Pegasus, 'anderson':Anderson,
              'ridder':Ridder, 'anewton':ANewton, 'mdnewton':MDNewton}

def findroot(ctx, f, x0, solver='secant', tol=None, verbose=False, verify=True, **kwargs):
    r"""
    Find a solution to `f(x) = 0`, using *x0* as starting point or
    interval for *x*.

    Multidimensional overdetermined systems are supported.
    You can specify them using a function or a list of functions.

    If the found root does not satisfy `|f(x)|^2 \leq \mathrm{tol}`,
    an exception is raised (this can be disabled with *verify=False*).

    **Arguments**

    *f*
        one dimensional function
    *x0*
        starting point, several starting points or interval (depends on solver)
    *tol*
        the returned solution has an error smaller than this
    *verbose*
        print additional information for each iteration if true
    *verify*
        verify the solution and raise a ValueError if `|f(x)|^2 > \mathrm{tol}`
    *solver*
        a generator for *f* and *x0* returning approximative solution and error
    *maxsteps*
        after how many steps the solver will cancel
    *df*
        first derivative of *f* (used by some solvers)
    *d2f*
        second derivative of *f* (used by some solvers)
    *multidimensional*
        force multidimensional solving
    *J*
        Jacobian matrix of *f* (used by multidimensional solvers)
    *norm*
        used vector norm (used by multidimensional solvers)

    solver has to be callable with ``(f, x0, **kwargs)`` and return an generator
    yielding pairs of approximative solution and estimated error (which is
    expected to be positive).
    You can use the following string aliases:
    'secant', 'mnewton', 'halley', 'muller', 'illinois', 'pegasus', 'anderson',
    'ridder', 'anewton', 'bisect'

    See mpmath.calculus.optimization for their documentation.

    **Examples**

    The function :func:`~mpmath.findroot` locates a root of a given function using the
    secant method by default. A simple example use of the secant method is to
    compute `\pi` as the root of `\sin x` closest to `x_0 = 3`::

        >>> from mpmath import *
        >>> mp.dps = 30; mp.pretty = True
        >>> findroot(sin, 3)
        3.14159265358979323846264338328

    The secant method can be used to find complex roots of analytic functions,
    although it must in that case generally be given a nonreal starting value
    (or else it will never leave the real line)::

        >>> mp.dps = 15
        >>> findroot(lambda x: x**3 + 2*x + 1, j)
        (0.226698825758202 + 1.46771150871022j)

    A nice application is to compute nontrivial roots of the Riemann zeta
    function with many digits (good initial values are needed for convergence)::

        >>> mp.dps = 30
        >>> findroot(zeta, 0.5+14j)
        (0.5 + 14.1347251417346937904572519836j)

    The secant method can also be used as an optimization algorithm, by passing
    it a derivative of a function. The following example locates the positive
    minimum of the gamma function::

        >>> mp.dps = 20
        >>> findroot(lambda x: diff(gamma, x), 1)
        1.4616321449683623413

    Finally, a useful application is to compute inverse functions, such as the
    Lambert W function which is the inverse of `w e^w`, given the first
    term of the solution's asymptotic expansion as the initial value. In basic
    cases, this gives identical results to mpmath's built-in ``lambertw``
    function::

        >>> def lambert(x):
        ...     return findroot(lambda w: w*exp(w) - x, log(1+x))
        ...
        >>> mp.dps = 15
        >>> lambert(1); lambertw(1)
        0.567143290409784
        0.567143290409784
        >>> lambert(1000); lambert(1000)
        5.2496028524016
        5.2496028524016

    Multidimensional functions are also supported::

        >>> f = [lambda x1, x2: x1**2 + x2,
        ...      lambda x1, x2: 5*x1**2 - 3*x1 + 2*x2 - 3]
        >>> findroot(f, (0, 0))
        [-0.618033988749895]
        [-0.381966011250105]
        >>> findroot(f, (10, 10))
        [ 1.61803398874989]
        [-2.61803398874989]

    You can verify this by solving the system manually.

    Please note that the following (more general) syntax also works::

        >>> def f(x1, x2):
        ...     return x1**2 + x2, 5*x1**2 - 3*x1 + 2*x2 - 3
        ...
        >>> findroot(f, (0, 0))
        [-0.618033988749895]
        [-0.381966011250105]


    **Multiple roots**

    For multiple roots all methods of the Newtonian family (including secant)
    converge slowly. Consider this example::

        >>> f = lambda x: (x - 1)**99
        >>> findroot(f, 0.9, verify=False)
        0.918073542444929

    Even for a very close starting point the secant method converges very
    slowly. Use ``verbose=True`` to illustrate this.

    It is possible to modify Newton's method to make it converge regardless of
    the root's multiplicity::

        >>> findroot(f, -10, solver='mnewton')
        1.0

    This variant uses the first and second derivative of the function, which is
    not very efficient.

    Alternatively you can use an experimental Newtonian solver that keeps track
    of the speed of convergence and accelerates it using Steffensen's method if
    necessary::

        >>> findroot(f, -10, solver='anewton', verbose=True)
        x:     -9.88888888888888888889
        error: 0.111111111111111111111
        converging slowly
        x:     -9.77890011223344556678
        error: 0.10998877665544332211
        converging slowly
        x:     -9.67002233332199662166
        error: 0.108877778911448945119
        converging slowly
        accelerating convergence
        x:     -9.5622443299551077669
        error: 0.107778003366888854764
        converging slowly
        x:     0.99999999999999999214
        error: 10.562244329955107759
        x:     1.0
        error: 7.8598304758094664213e-18
        ZeroDivisionError: canceled with x = 1.0
        1.0

    **Complex roots**

    For complex roots it's recommended to use Muller's method as it converges
    even for real starting points very fast::

        >>> findroot(lambda x: x**4 + x + 1, (0, 1, 2), solver='muller')
        (0.727136084491197 + 0.934099289460529j)


    **Intersection methods**

    When you need to find a root in a known interval, it's highly recommended to
    use an intersection-based solver like ``'anderson'`` or ``'ridder'``.
    Usually they converge faster and more reliable. They have however problems
    with multiple roots and usually need a sign change to find a root::

        >>> findroot(lambda x: x**3, (-1, 1), solver='anderson')
        0.0

    Be careful with symmetric functions::

        >>> findroot(lambda x: x**2, (-1, 1), solver='anderson') #doctest:+ELLIPSIS
        Traceback (most recent call last):
          ...
        ZeroDivisionError

    It fails even for better starting points, because there is no sign change::

        >>> findroot(lambda x: x**2, (-1, .5), solver='anderson')
        Traceback (most recent call last):
          ...
        ValueError: Could not find root within given tolerance. (1.0 > 2.16840434497100886801e-19)
        Try another starting point or tweak arguments.

    """
    prec = ctx.prec
    try:
        ctx.prec += 20

        # initialize arguments
        if tol is None:
            tol = ctx.eps * 2**10

        kwargs['verbose'] = kwargs.get('verbose', verbose)

        if 'd1f' in kwargs:
            kwargs['df'] = kwargs['d1f']

        kwargs['tol'] = tol
        if isinstance(x0, (list, tuple)):
            x0 = [ctx.convert(x) for x in x0]
        else:
            x0 = [ctx.convert(x0)]

        if isinstance(solver, str):
            try:
                solver = str2solver[solver]
            except KeyError:
                raise ValueError('could not recognize solver')

        # accept list of functions
        if isinstance(f, (list, tuple)):
            f2 = copy(f)
            def tmp(*args):
                return [fn(*args) for fn in f2]
            f = tmp

        # detect multidimensional functions
        try:
            fx = f(*x0)
            multidimensional = isinstance(fx, (list, tuple, ctx.matrix))
        except TypeError:
            fx = f(x0[0])
            multidimensional = False
        if 'multidimensional' in kwargs:
            multidimensional = kwargs['multidimensional']
        if multidimensional:
            # only one multidimensional solver available at the moment
            solver = MDNewton
            if not 'norm' in kwargs:
                norm = lambda x: ctx.norm(x, 'inf')
                kwargs['norm'] = norm
            else:
                norm = kwargs['norm']
        else:
            norm = abs

        # happily return starting point if it's a root
        if norm(fx) == 0:
            if multidimensional:
                return ctx.matrix(x0)
            else:
                return x0[0]

        # use solver
        iterations = solver(ctx, f, x0, **kwargs)
        if 'maxsteps' in kwargs:
            maxsteps = kwargs['maxsteps']
        else:
            maxsteps = iterations.maxsteps
        i = 0
        for x, error in iterations:
            if verbose:
                print_('x:    ', x)
                print_('error:', error)
            i += 1
            if error < tol * max(1, norm(x)) or i >= maxsteps:
                break
        else:
            if not i:
                raise ValueError('Could not find root using the given solver.\n'
                                 'Try another starting point or tweak arguments.')
        if not isinstance(x, (list, tuple, ctx.matrix)):
            xl = [x]
        else:
            xl = x
        if verify and norm(f(*xl))**2 > tol: # TODO: better condition?
            raise ValueError('Could not find root within given tolerance. '
                             '(%s > %s)\n'
                             'Try another starting point or tweak arguments.'
                             % (norm(f(*xl))**2, tol))
        return x
    finally:
        ctx.prec = prec


def multiplicity(ctx, f, root, tol=None, maxsteps=10, **kwargs):
    """
    Return the multiplicity of a given root of f.

    Internally, numerical derivatives are used. This might be inefficient for
    higher order derviatives. Due to this, ``multiplicity`` cancels after
    evaluating 10 derivatives by default. You can be specify the n-th derivative
    using the dnf keyword.

    >>> from mpmath import *
    >>> multiplicity(lambda x: sin(x) - 1, pi/2)
    2

    """
    if tol is None:
        tol = ctx.eps ** 0.8
    kwargs['d0f'] = f
    for i in xrange(maxsteps):
        dfstr = 'd' + str(i) + 'f'
        if dfstr in kwargs:
            df = kwargs[dfstr]
        else:
            df = lambda x: ctx.diff(f, x, i)
        if not abs(df(root)) < tol:
            break
    return i

def steffensen(f):
    """
    linear convergent function -> quadratic convergent function

    Steffensen's method for quadratic convergence of a linear converging
    sequence.
    Don not use it for higher rates of convergence.
    It may even work for divergent sequences.

    Definition:
    F(x) = (x*f(f(x)) - f(x)**2) / (f(f(x)) - 2*f(x) + x)

    Example
    .......

    You can use Steffensen's method to accelerate a fixpoint iteration of linear
    (or less) convergence.

    x* is a fixpoint of the iteration x_{k+1} = phi(x_k) if x* = phi(x*). For
    phi(x) = x**2 there are two fixpoints: 0 and 1.

    Let's try Steffensen's method:

    >>> f = lambda x: x**2
    >>> from mpmath.calculus.optimization import steffensen
    >>> F = steffensen(f)
    >>> for x in [0.5, 0.9, 2.0]:
    ...     fx = Fx = x
    ...     for i in xrange(9):
    ...         try:
    ...             fx = f(fx)
    ...         except OverflowError:
    ...             pass
    ...         try:
    ...             Fx = F(Fx)
    ...         except ZeroDivisionError:
    ...             pass
    ...         print('%20g  %20g' % (fx, Fx))
                    0.25                  -0.5
                  0.0625                   0.1
              0.00390625            -0.0011236
             1.52588e-05           1.41691e-09
             2.32831e-10          -2.84465e-27
             5.42101e-20           2.30189e-80
             2.93874e-39          -1.2197e-239
             8.63617e-78                     0
            7.45834e-155                     0
                    0.81               1.02676
                  0.6561               1.00134
                0.430467                     1
                0.185302                     1
               0.0343368                     1
              0.00117902                     1
             1.39008e-06                     1
             1.93233e-12                     1
             3.73392e-24                     1
                       4                   1.6
                      16                1.2962
                     256               1.10194
                   65536               1.01659
             4.29497e+09               1.00053
             1.84467e+19                     1
             3.40282e+38                     1
             1.15792e+77                     1
            1.34078e+154                     1

    Unmodified, the iteration converges only towards 0. Modified it converges
    not only much faster, it converges even to the repelling fixpoint 1.
    """
    def F(x):
        fx = f(x)
        ffx = f(fx)
        return (x*ffx - fx**2) / (ffx - 2*fx + x)
    return F

OptimizationMethods.jacobian = jacobian
OptimizationMethods.findroot = findroot
OptimizationMethods.multiplicity = multiplicity

if __name__ == '__main__':
    import doctest
    doctest.testmod()